Asked by Megan

A 20.0 g bullet is fired horizontally into a 85 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 156 N/m. The bullet becomes embedded in the block. The bullet block system compresses the spring by a maximum of 1.30 cm. (a) What is the velocity of the block once the bullet is imbedded? (b)What was the speed of the bullet at impact with the block?
Answered by Elena
Law of conservation of energy
KE =PE
(m1+m2) •u²/2 =kx²/2
u=sqrt{kx²/(m1+m2)}=.....
Law of conservation og linear momentum
m1v + 0= (m1+m2)u
v=(m1+m2)u/m1 =....
Answered by Richmond
Work backwards, because you cannot solve it forwards since there are two unknowns.

(a) Start with the energy transferred to the compressed spring. The block transfers energy to the spring, that's why it compresses.

Ek = Ee
mv^2 = kx^2 (halves cancel)
v = sqrt[(kx^2)/m]
v = .310 m/s

(b) Momentum is conserved. The momentum before is the momentum of the bullet. After, it is the combined mass of bullet and block.
Let u be the initial velocity of bullet
mu = (m1+m2)v
u = 1.63 m/s
Answered by Kris
RICHMOND- you are sir should stop trying to show people how to do physics when you are completely wrong.
There are no AI answers yet. The ability to request AI answers is coming soon!